GOlang：将 XML 嵌套到 JSON

我试图弄清楚如何给出 XML 输入，使用 GOlang 将其转换为 JSON。例如...

<version>0.1</version>

    <termsofService>http://www.wunderground.com/weather/api/d/terms.html</termsofService>

    <features>

        <feature>conditions</feature>

    </features>

会变成

"version": "0.1",

    "termsofService": "http://www.wunderground.com/weather/api/d/terms.html",

    "features": { "feature": "conditions" },

我得到了versionandtermsofservice正确，但我不知道如何返回嵌套的features. 这是我必须硬编码的东西吗？

代码：

    type reportType struct{

    Version xml.CharData        `xml:"version"`

    TermsOfService xml.CharData `xml:"termsofService"

    `

    Features xml.CharData       `xml:"features>feature"`

    Zip      xml.CharData       `xml:"current_observation>display_location>zip"`

    Problem myErrorType     `xml:"error"`

}

type myErrorType struct{

    TypeOfError xml.CharData `xml:"type"`

    Desciption xml.CharData `xml:"description"`

}

type reportTypeJson struct{

    Version        string  `json:"version"`;

    TermsOfService string `json:"termsofService"`;

    Features    string `json:"features feature" `;

    Zip           string `json:"current_observation > display_location > zip"`;

}

func main() {

    fmt.Println("data is from WeatherUnderground.")

    fmt.Println("https://www.wunderground.com/")

    var state, city string

    str1 := "What is your state?"

    str2 := "What is your city?"

    fmt.Println(str1)

    fmt.Scanf("%s", &state)

    fmt.Println(str2)

    fmt.Scanf("%s", &city)

    baseURL := "http://api.wunderground.com/api/";

    apiKey := "YouDontNeedToKnow"

    var query string

    //set up the query

    query = baseURL+apiKey +

    "/conditions/q/"+

    url.QueryEscape(state)+ "/"+

    url.QueryEscape(city)+ ".xml"

}

输出：

JSON output nicely formatted: 

{

      "version": "0.1",

      "termsofService": "http://www.wunderground.com/weather/api/d/terms.html",

      "features \u003e feature": "conditions",

      "current_observation \u003e display_location \u003e zip": "64068"

}

谢谢你的时间！